Question: What is the area of the region bound by the graphs of $f(x)=\sqrt{x-2}$, $g(x)=14-x$, and $x=2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{99}{2}$ (Choice B) B $\dfrac{19}{6}$ (Choice C) C $\dfrac{45}{2}$ (Choice D) D $\dfrac{151}{2}$
Solution: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${2}$ ${4}$ ${6}$ ${8}$ ${10}$ ${12}$ ${14}$ ${2}$ ${4}$ ${6}$ ${8}$ ${10}$ ${12}$ ${14}$ $f$ $g$ $y$ $x$ From the graph, it appears that $g(x)\ge f(x)$ between $x=2$ and the point where the graphs intersect. From this we are looking to evaluate: $ \int_{2}^{b}\left( g(x)-f(x) \right)\,dx$ where $b$ is the $x$ -coordinate of the point of intersection. Finding the $x$ -coordinate of the intersection point We can find the $x$ -coordinate of the point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ \sqrt{x-2}&=14-x\\\\ x-2 &=x^2-28x+196 \\\\ 0 &= x^2-29x+198 \\\\ 0 &= (x-11) (x-18) \end{aligned}$ We have two possible solutions at $x={11}$ and $x={18}$. Because this was a radical equation, let's check for extraneous solutions by plugging our possible solutions back into the original equation. $\begin{aligned} \sqrt{{11}-2}&\stackrel{?}=14-{11}\\\\ 3 & \stackrel{\checkmark}= 3 \end{aligned}$ $\begin{aligned} \sqrt{{18}-2}&\stackrel{?}=14-{18}\\\\ 4 & \neq -4 \end{aligned}$ The graphs intersect where $x=11$. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $ \int_{2}^{11}\left(14-x-(\sqrt{x-2})\right)\,dx$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{2}^{11} \left( 14-x- (\sqrt{x-2}) \right) \,dx \\\\ &= 14x-\dfrac{x^2}{2}-\dfrac{2(x-2)^{3/2}}{3}~\Bigg|_{2}^{11} \\\\ &= \left( 154-\dfrac{121}{2}-18 \right) - \left( 28-2-0 \right) \\\\ &= \dfrac{99}{2} \end{aligned}$ Answer The area is $\dfrac{99}{2}$ square units.